The two possible angles obtained by using the qudratic equation are;
θ
= 15.10° and θ2 = 73.51°
Given, speed of water = 
= 50ft/s
For the motion along x direction, time period can be calculated as follows:
35 = (50 × cosθ) t
t = 0.64 / cosθ
For the motion in y direction, an equation can be obtained as follows:
θ) 
Plugging in the values we get:
θ) 
-20 = -32tanθ - 10.304
θ
Upon solving the above quadratic equation, we get,
tanθ = 0.27 , -3.38
Therefore,
tanθ = 0.27
θ = 15.10°
and, tanθ
= -3.38
θ = 73.51
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Dew point is the temperature at which the water vapor in the air condenses , then evaporates. The barometric or air pressure is independent from the dew point.
Answer:
M[min] = M[basket+people+ balloon, not gas] * ΔR/R[b]
ΔR is the difference in density between the gas inside and surrounding the balloon.
R[b] is the density of gas inside the baloon.
====================================
Let V be the volume of helium required.
Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V
U = 1.225gV newtons
----
Weight of Helium = Volume of Helium * Density of Helium * g
W[h] = 0.18gV N
Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N -----
Weight of 260kg = 2549.7 N
Then to lift the whole thing, F > 2549.7
So minimal F would be 2549.7
----
1.045gV = 2549.7
V = 248.8 m^3
Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf)
=====
Let the density of the surroundings be R
Then U-W = (1-0.9)RgV = 0.1RgV
So 0.1RgV = 2549.7 N
V = 2549.7 / 0.1Rg
Assuming that R is again 1.255, V = 2071.7 m^3
Then mass of hot air required = 230.2 * 0.9R = 2340 kg
Notice from this that M = 2549.7/0.9Rg * 0.1R so
M[min] = Weight of basket * (difference in density between balloon's gas and surroundings / density of gas in balloon)
M[min] = M[basket] * ΔR/R[b]
